November 15, 2019

Solution Set 5

Physics 208-02 – Introductory Physics II
Solution Set 5
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
Spring 2012

§ Chapter 12, Problem 7

Calculate the percent error made over a distance of one mile by the “five-second rule” for estimating the distance of a lightning strike for the temperatures indicated. (The rule states a five-second delay between the flash of lightning and the ensuing sound of thunder for every mile from the strike treating the delay for the travel of light as negligible.)

§ (a)


Compute the speed of sound at 30^{0}\;\mathrm{C}.

\displaystyle v \displaystyle= \displaystyle\left(331+\frac{0.60}{\mbox{}^{0}\mathrm{C}}30^{0}\;\mathrm{C}\right)\frac{\mathrm{m}}{\mathrm{s}}
\displaystyle= \displaystyle 349\frac{\mathrm{m}}{\mathrm{s}}

The distance traveled in five seconds is then

\displaystyle d \displaystyle= \displaystyle vt
\displaystyle= \displaystyle\left(349\frac{\mathrm{m}}{\mathrm{s}}\right)(5\;\mathrm{s})
\displaystyle= \displaystyle 1745\;\mathrm{m}\times\frac{1\;\mathrm{mile}}{1609\;\mathrm{m}}
\displaystyle= \displaystyle 1.08\;\mathrm{mile}

The error is then approximately

\displaystyle\mathrm{Percent\; Error} \displaystyle= \displaystyle\frac{1.08\;\mathrm{mile}-1\;\mathrm{mile}}{1\;\mathrm{mile}}\times 100
\displaystyle= \displaystyle 8\;\mathrm{percent}

§ (b)


Compute the speed of sound at 10^{0}\;\mathrm{C}.

\displaystyle v \displaystyle= \displaystyle\left(331+\frac{0.60}{\mbox{}^{0}\mathrm{C}}10^{0}\;\mathrm{C}\right)\frac{\mathrm{m}}{\mathrm{s}}
\displaystyle= \displaystyle 337\frac{\mathrm{m}}{\mathrm{s}}

The distance traveled in five seconds is then

\displaystyle d \displaystyle= \displaystyle vt
\displaystyle= \displaystyle\left(337\frac{\mathrm{m}}{\mathrm{s}}\right)(5\;\mathrm{s})
\displaystyle= \displaystyle 1685\;\mathrm{m}\times\frac{1\;\mathrm{mile}}{1609\;\mathrm{m}}
\displaystyle= \displaystyle 1.05\;\mathrm{mile}

The error is then approximately

\displaystyle\mathrm{Percent\; Error} \displaystyle= \displaystyle\frac{1.05\;\mathrm{mile}-1\;\mathrm{mile}}{1\;\mathrm{mile}}\times 100
\displaystyle= \displaystyle 5\;\mathrm{percent}

§ Chapter 12, Problem 10

If two firecrackers produce a sound level of 95\;\mathrm{dB} when fired simultaneously at a certain place, what will the sound level be if only one is fired? (Hint, intensities add, not sound levels).

\displaystyle 95\;\mathrm{dB} \displaystyle= \displaystyle 10\;\mathrm{dB}\,\log\left(\frac{I_{2}}{1.0\times 10^{{-12}}\frac{\mathrm{W}}{\mathrm{m}^{2}}}\right)
\displaystyle I_{2} \displaystyle= \displaystyle\left(1.0\times 10^{{-12}}\frac{\mathrm{W}}{\mathrm{m}^{2}}\right)10^{{(95\;\mathrm{dB}/10\;\mathrm{dB})}}
\displaystyle= \displaystyle 3.\underline{1}6\times 10^{{-3}}\frac{\mathrm{W}}{\mathrm{m}^{2}}
\displaystyle I_{1} \displaystyle= \displaystyle\frac{1}{2}I_{2}
\displaystyle= \displaystyle 1.\underline{5}8\times 10^{{-3}}\frac{\mathrm{W}}{\mathrm{m}^{2}}
\displaystyle\mathrm{level} \displaystyle= \displaystyle 10\;\mathrm{dB}\,\log\left(\frac{1.\underline{5}8\times 10^{{-3}}\frac{\mathrm{W}}{\mathrm{m}^{2}}}{1.0\times 10^{{-12}}\frac{\mathrm{W}}{\mathrm{m}^{2}}}\right)
\displaystyle= \displaystyle 92\;\mathrm{dB}

There is a shortcut here. Note the following.

\displaystyle 10\;\mathrm{dB}\,\log\left(\frac{I/2}{I_{0}}\right) \displaystyle= \displaystyle 10\;\mathrm{dB}\,\log\left(\frac{1}{2}\frac{I}{I_{0}}\right)
\displaystyle= \displaystyle 10\;\mathrm{dB}\,\log\left(\frac{I}{I_{0}}\right)-10\;\mathrm{dB}\,\log\left(2\right)
\displaystyle= \displaystyle 10\;\mathrm{dB}\,\log\left(\frac{I}{I_{0}}\right)-3.0

§ Chapter 12, Problem 12

A casette player is said to have a signal to noise ratio of 58\;\mathrm{dB}, whereas for a CD player it is 95\;\mathrm{dB}. What is the ration of intensities for each device?

For the cassette player:

\displaystyle\frac{I}{I_{0}} \displaystyle= \displaystyle 10^{{58\;\mathrm{dB}/10\;\mathrm{dB}}}
\displaystyle= \displaystyle 6.3\times 10^{5}

For the CD player:

\displaystyle\frac{I}{I_{0}} \displaystyle= \displaystyle 10^{{95\;\mathrm{dB}/10\;\mathrm{dB}}}
\displaystyle= \displaystyle 3.2\times 10^{9}

That’s an improvement by a factor of 5,000 in terms of the intensity ratio!

§ Chapter 12, Problem 14

A 50\;\mathrm{dB} sound wave strikes an eardrum whose area is 5.0\times 10^{{-5}}\mathrm{m}^{2}.

§ 0.1. (a)

How much energy is absorbed by the eardrum per second?

Begin by computing the intensity which is power per unit area:

\displaystyle 50\;\mathrm{dB} \displaystyle= \displaystyle 10\;\mathrm{dB}\,\log\left(\frac{I}{I_{0}}\right)
\displaystyle I \displaystyle= \displaystyle I_{0}\, 10^{{5}}
\displaystyle= \displaystyle\left(1.0\times 10^{{-12}}\frac{\mathrm{W}}{\mathrm{m}^{2}}\right)\left(10^{5}\right)
\displaystyle= \displaystyle 1.0\times 10^{{-7}}\frac{\mathrm{W}}{\mathrm{m}^{2}}

Now compute the power by multiplying by the area:

\displaystyle P \displaystyle= \displaystyle\left(1.0\times 10^{{-7}}\frac{\mathrm{W}}{\mathrm{m}^{2}}\right)\left(5.0\times 10^{{-5}}\mathrm{m}^{2}\right)
\displaystyle= \displaystyle 5.0\times 10^{{-12}}\;\mathrm{W}
\displaystyle= \displaystyle 5.0\times 10^{{-12}}\frac{\mathrm{J}}{\mathrm{s}}

Notice the units. Power is energy per unit time.

§ 0.2. (b)

How long would it take your eardrum to receive a total energy of 1.0\;\mathrm{J}?

If you don’t remember the equations, just pay attention to the units.

\displaystyle E \displaystyle= \displaystyle Pt
\displaystyle t \displaystyle= \displaystyle\frac{E}{P}
\displaystyle= \displaystyle\frac{1.0\;\mathrm{J}}{5.0\times 10^{{-12}}\frac{\mathrm{J}}{\mathrm{s}}}
\displaystyle= \displaystyle 2.0\times 10^{{11}}\mathrm{s}\times\frac{1\;\mathrm{hr}}{3600\;\mathrm{s}}\times\frac{1\;\mathrm{day}}{24\;\mathrm{hr}}\times\frac{1\;\mathrm{year}}{365.25\;\mathrm{day}}
\displaystyle= \displaystyle 6,300\;\mathrm{year}

§ Chapter 12, Problem 15

Expensive amplifier A is rated at 250\;\mathrm{W} while the more modest amplifier B is rated at 40\;\mathrm{W}.

§ (a)

Estimate the sound level in decibels that you would expect at a point 3.5\;\mathrm{m} from a loudspeaker connected in turn to each amplifier.

While loudspeakers are somewhat directional, our best estimate without more information is that the energy is distributed over a spherical surface. For amplifier A:

\displaystyle I \displaystyle= \displaystyle\frac{250\;\mathrm{W}}{4\pi\left(3.5\;\mathrm{m}\right)^{2}}
\displaystyle= \displaystyle 1.\underline{6}2\frac{\mathrm{W}}{\mathrm{m}^{2}}
\displaystyle\beta \displaystyle= \displaystyle 10\;\mathrm{dB}\,\log\left(\frac{1.\underline{6}2\frac{\mathrm{W}}{\mathrm{m}^{2}}}{1.0\times 10^{{-12}}\frac{\mathrm{W}}{\mathrm{m}^{2}}}\right)
\displaystyle= \displaystyle 1\underline{2}2\;\mathrm{dB}

For amplifier B:

\displaystyle I \displaystyle= \displaystyle\frac{40\;\mathrm{W}}{4\pi\left(3.5\;\mathrm{m}\right)^{2}}
\displaystyle= \displaystyle 0.\underline{2}6\frac{\mathrm{W}}{\mathrm{m}^{2}}
\displaystyle\beta \displaystyle= \displaystyle 10\;\mathrm{dB}\,\log\left(\frac{0.\underline{2}6\frac{\mathrm{W}}{\mathrm{m}^{2}}}{1.0\times 10^{{-12}}\frac{\mathrm{W}}{\mathrm{m}^{2}}}\right)
\displaystyle= \displaystyle\underline{1}14\;\mathrm{dB}

§ (b)

Will the expensive amp sound twice as loud as the cheaper one?

No, it will not. It will sound louder by approximately

\displaystyle\frac{1\underline{2}2\;\mathrm{dB}-\underline{1}14\;\mathrm{dB}}{\underline{1}14\;\mathrm{dB}}\times 100\;\mathrm{percent} \displaystyle= \displaystyle 7\;\mathrm{percent}

§ Chapter 12, Problem 24

A string on a violin has a fundamental frequency of 440\;\mathrm{Hz}. The length of the vibrating portion is 32\;\mathrm{cm} with a mass of 0.35\;\mathrm{g}. Under what tension must the string be placed?

With the wave speed given by

\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{T}{m/L}}

and the fundamental frequency given by

\displaystyle f_{1} \displaystyle= \displaystyle\frac{v}{2L}
\displaystyle v \displaystyle= \displaystyle 2Lf_{1}

we may write

\displaystyle\sqrt{\frac{T}{m/L}} \displaystyle= \displaystyle 2Lf_{1}
\displaystyle\frac{T}{m/L} \displaystyle= \displaystyle 4L^{2}f_{1}^{2}
\displaystyle T \displaystyle= \displaystyle 4mLf_{1}^{2}
\displaystyle= \displaystyle 4\left(3.5\times 10^{{-4}}\mathrm{kg}\right)\left(3.2\times 10^{{-1}}\;\mathrm{m}\right)\left(400\;\mathrm{s}^{{-1}}\right)^{2}
\displaystyle= \displaystyle 72\;\mathrm{N}

§ Chapter 12, Problem 39

A piano tuner hears one beat every 2.0\;\mathrm{s} when trying to adjust two strings, one of which is sounding 440\;\mathrm{Hz}. How far off in frequency is the other string?

We are given the beat period, so use it to calculate the beat frequency.

\displaystyle T_{\mathrm{beat}} \displaystyle= \displaystyle 2.0\;\mathrm{s}
\displaystyle f_{\mathrm{beat}} \displaystyle= \displaystyle T_{\mathrm{beat}}^{{-1}}
\displaystyle= \displaystyle\left(2.0\;\mathrm{s}\right)^{{-1}}
\displaystyle= \displaystyle 0.50\;\mathrm{Hz}

So the frequency of the second string differs from the first by 0.50\;\mathrm{Hz}. In other words, it is either 440.5\;\mathrm{Hz} or 339.5\;\mathrm{Hz}.

§ Chapter 12, Problem 49

The predominant frequency of a certain fire engine’s siren is 1550\;\mathrm{Hz} when at rest.

§ (a)

What frequency do you detect when you move 30.0\;\mathrm{m}/\mathrm{s} toward the fire engine?

The equation for the Doppler shift is given by

\displaystyle f^{\prime} \displaystyle= \displaystyle f\frac{v_{\mathrm{sound}}\pm v_{\mathrm{observer}}}{v_{\mathrm{sound}}\mp v_{\mathrm{source}}}

In this case the speed of the source is zero, so

\displaystyle f^{\prime} \displaystyle= \displaystyle f\frac{v_{\mathrm{sound}}\pm v_{\mathrm{observer}}}{v_{\mathrm{sound}}}

Moving towards the source we anticipate the pitch will increase, so choose the addition rather than subtraction where there is the \pm symbol. Then

\displaystyle f^{\prime} \displaystyle= \displaystyle 1550\;\mathrm{Hz}\frac{331\frac{\mathrm{m}}{\mathrm{s}}+30.0\frac{\mathrm{m}}{\mathrm{s}}}{331\frac{\mathrm{m}}{\mathrm{s}}}
\displaystyle= \displaystyle 1\underline{6}90\;\mathrm{Hz}

§ (b)

What frequency do you detect when you move 30.0\;\mathrm{m}/\mathrm{s} away from the fire engine?

Follow exactly the same logic as part (a) with the change that \pm\rightarrow- because we expect the pitch to drop when moving away from the source. Then

\displaystyle f^{\prime} \displaystyle= \displaystyle 1550\;\mathrm{Hz}\frac{331\frac{\mathrm{m}}{\mathrm{s}}-30.0\frac{\mathrm{m}}{\mathrm{s}}}{331\frac{\mathrm{m}}{\mathrm{s}}}
\displaystyle= \displaystyle 1\underline{4}10\;\mathrm{Hz}

§ Chapter 12, Problem 53

A bat at rest send out ultrasonic sound waves at 50.0\;\mathrm{kHz} and receives them returned from an object moving directly away from it at 25.0\;\mathrm{m}/\mathrm{s}. What is the received frequency?

Start with the equation for the Doppler shift given by

\displaystyle f^{\prime} \displaystyle= \displaystyle f\frac{v_{\mathrm{sound}}\pm v_{\mathrm{observer}}}{v_{\mathrm{sound}}\mp v_{\mathrm{source}}}

First the object acts as the observer moving away at v_{\mathrm{observer}}=25.0\;\mathrm{m}/\mathrm{s}, and we choose \pm\rightarrow- in the numerator because the pitch will drop. The object now acts as a source with v_{\mathrm{source}}=25.0\;\mathrm{m}/\mathrm{s}, and we choose \mp\rightarrow+ because the pitch will drop with the source moving away. Hence

\displaystyle f^{\prime} \displaystyle= \displaystyle 50.0\;\mathrm{kHz}\frac{331\frac{\mathrm{m}}{\mathrm{s}}-25.0\frac{\mathrm{m}}{\mathrm{s}}}{331\frac{\mathrm{m}}{\mathrm{s}}+25.0\frac{\mathrm{m}}{\mathrm{s}}}
\displaystyle= \displaystyle 43\;\mathrm{kHz}

§ Chapter 12, Problem 55

In one of the original Doppler experiments a tuba was played on a moving flat train car at a frequency of 75\;\mathrm{Hz}, and an identical tuba at rest in the station played the same tone. What beat frequency was heard as the train approached the station at 10.0\;\mathrm{m}/\mathrm{s}?

First compute the Doppler shifted frequency with a moving source choosing subtraction in the denominator for increased pitch.

\displaystyle f^{\prime} \displaystyle= \displaystyle f\frac{v_{\mathrm{sound}}}{v_{\mathrm{sound}}-v_{\mathrm{source}}}
\displaystyle= \displaystyle\left(75\;\mathrm{Hz}\right)\frac{331\frac{\mathrm{m}}{\mathrm{s}}}{331\frac{\mathrm{m}}{\mathrm{s}}-10\frac{\mathrm{m}}{\mathrm{s}}}
\displaystyle= \displaystyle 77\;\mathrm{Hz}


\displaystyle f_{\mathrm{beat}} \displaystyle= \displaystyle\left|77\;\mathrm{Hz}-75\;\mathrm{Hz}\right|
\displaystyle= \displaystyle 2\;\mathrm{Hz}


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